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Difficulty: HardCoordinate Geometry and Lines

In the xyxy-plane, line LL passes through the point (4,2)(4, 2) and has a negative slope mm. Line LL intersects the positive xx-axis at point PP and the positive yy-axis at point QQ. If the area of triangle OPQOPQ, where OO is the origin, is equal to 1616, what is the slope of line LL?

  1. A
    −2-2
  2. B
    −1-1
  3. −12-\frac{1}{2}Answer
  4. D
    −14-\frac{1}{4}
  5. E
    12\frac{1}{2}

Answer

The slope of line LL is −12-\frac{1}{2}.
The line equation in point-slope form is y−2=m(x−4)y - 2 = m(x - 4). Setting y=0y = 0 yields the xx-intercept P(4−2m,0)P\left(4 - \frac{2}{m}, 0\right), and setting x=0x = 0 yields the yy-intercept Q(0,2−4m)Q(0, 2 - 4m). The area of right triangle OPQOPQ is 12(4−2m)(2−4m)=16\frac{1}{2}\left(4 - \frac{2}{m}\right)(2 - 4m) = 16. Expanding this equation leads to 4m2+4m+1=04m^2 + 4m + 1 = 0, which factors as (2m+1)2=0(2m + 1)^2 = 0, giving m=−12m = -\frac{1}{2}.

Step-by-Step Solution

1
Write the point-slope form of line LL using point (4,2)(4, 2) and slope mm.
The equation of line LL is y−2=m(x−4)y - 2 = m(x - 4).
Point-slope form allows expressing intercepts directly in terms of the unknown slope mm.
2
Find the coordinates of the xx-intercept PP and yy-intercept QQ.
Setting y=0y = 0 gives xP=4−2mx_P = 4 - \frac{2}{m}. Setting x=0x = 0 gives yQ=2−4my_Q = 2 - 4m.
Intersections with the axes define the base and height of right triangle OPQOPQ.
3
Set up the area equation for triangle OPQOPQ.
\text{Area} = \frac{1}{2} \cdot x_P \cdot y_Q = \frac{1}{2} \left(4 - \frac{2}{m}\right)(2 - 4m) = 16.
Triangle OPQOPQ is a right triangle with base along the xx-axis and height along the yy-axis.
4
Solve the algebraic equation for slope mm.
Expanding gives 12(8−16m−4m+8)=16  ⟹  16−16m−4m=32  ⟹  −16m−16−4m=0\frac{1}{2}\left(8 - 16m - \frac{4}{m} + 8\right) = 16 \implies 16 - 16m - \frac{4}{m} = 32 \implies -16m - 16 - \frac{4}{m} = 0. Dividing by −4-4 and multiplying by mm yields 4m2+4m+1=0  ⟹  (2m+1)2=0  ⟹  m=−124m^2 + 4m + 1 = 0 \implies (2m + 1)^2 = 0 \implies m = -\frac{1}{2}.
Factoring the perfect square quadratic equation yields the single real value for mm.

Key Concept

Coordinate Geometry: Intercepts and Area of Triangles

Alternative Method

Use the intercept form of a line xa+yb=1\frac{x}{a} + \frac{y}{b} = 1. The area condition gives ab=32ab = 32. Substituting (4,2)(4, 2) gives 4a+2b=1\frac{4}{a} + \frac{2}{b} = 1. Substituting b=32ab = \frac{32}{a} leads to 4a+2a32=1  ⟹  4a+a16=1  ⟹  a2−16a+64=0  ⟹  (a−8)2=0  ⟹  a=8\frac{4}{a} + \frac{2a}{32} = 1 \implies \frac{4}{a} + \frac{a}{16} = 1 \implies a^2 - 16a + 64 = 0 \implies (a-8)^2 = 0 \implies a = 8. Then b=4b = 4, and the slope is m=−ba=−48=−12m = -\frac{b}{a} = -\frac{4}{8} = -\frac{1}{2}.
Estimated Time:2m 0s
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Topics

Coordinate Geometry and Lines

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